Question involving a collection of closed intervals

**Runty** · September 19th 2010, 11:06 AM

Let $F$ be the collection of closed intervals $A_n=[1/n, 1-1/n]$ for $n=3,4,5,...$ . What do you notice about $\bigcup F$ ? Is it closed, open, both, or neither?

I'm afraid the wording and available material has got me stumped on this, and being sick at the moment doesn't help.

**Plato** · September 19th 2010, 11:19 AM

Notice that $\left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$ .
Can either 1 or 0 be included?

**Runty** · September 19th 2010, 11:43 AM

Originally Posted by Plato

Notice that $\left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$ .
Can either 1 or 0 be included?

I should note, the original question said that $n=1,2,3,...$ , but our Prof. told us to change it to $n=3,4,5,...$ when he realized that when $n=1, A_n=[1,0]$ and when $n=2, A_n=[1/2,1/2]$ (or at least that's what I heard him say). I wrote the question word-for-word (besides the change our Prof. gave us), so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.

**Plato** · September 19th 2010, 11:56 AM

Originally Posted by Runty

so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.

That is the whole point: neither 0 nor 1 can be included.
So what is the union?

**Runty** · September 19th 2010, 12:05 PM

Originally Posted by Plato

That is the whole point: neither 0 nor 1 can be included.
So what is the union?

I am presently guessing, but here is my answer thus far.

$\bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$

My guess so far indicates that the union is both open and closed, though I could easily be wrong.

**Plato** · September 19th 2010, 12:15 PM

Originally Posted by Runty

I am presently guessing, but here is my answer thus far.

$\bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$

For any $k$ does $1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
If not how can the union be $(0,1]?$

**Runty** · September 19th 2010, 12:16 PM

Originally Posted by Plato

For any $k$ does $1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
If not how can the union be $(0,1]?$

Point taken. I guess it's a completely open interval, then.

EDIT: One more thing, I've looked at some work done on this in another forum, for which I will provide the link.

http://www.mymathforum.com/viewtopic...=15805&start=0

Is the work found here accurate?

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